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thomas5267
 one year ago
Prove that the polynomials \(P_n\) has n distinct root for all n. \(P_n\) are characteristic polynomials of a particular type of matrix.
\[
P_n=xA_{n1}\frac{1}{2}A_{n2}\\
A_n = \left( \dfrac{x+\sqrt{x^21}}{2}\right)^n + \left(\dfrac{x\sqrt{x^21}}{2}\right)^n
\]
\(A_n\) satisfies the following recurrence relation:
\[
A_n=x A_{n1}\frac{1}{4}A_{n2}\\
A_1=x\\
A_2=x^2\frac{1}{2}
\]
thomas5267
 one year ago
Prove that the polynomials \(P_n\) has n distinct root for all n. \(P_n\) are characteristic polynomials of a particular type of matrix. \[ P_n=xA_{n1}\frac{1}{2}A_{n2}\\ A_n = \left( \dfrac{x+\sqrt{x^21}}{2}\right)^n + \left(\dfrac{x\sqrt{x^21}}{2}\right)^n \] \(A_n\) satisfies the following recurrence relation: \[ A_n=x A_{n1}\frac{1}{4}A_{n2}\\ A_1=x\\ A_2=x^2\frac{1}{2} \]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not sure if we need to find th characteristic coefficient and show they are distinct

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I want to prove these matrices are diagonalisable. \[ M_5=\begin{pmatrix} 0&\frac{1}{2}&0&0&0\\ 1&0&\frac{1}{2}&0&0\\ 0&\frac{1}{2}&0&\frac{1}{2}&0\\ 0&0&\frac{1}{2}&0&1\\ 0&0&0&\frac{1}{2}&0 \end{pmatrix}\\ M_7=\begin{pmatrix} 0&\frac{1}{2}&0&0&0&0&0\\ 1&0&\frac{1}{2}&0&0&0&0\\ 0&\frac{1}{2}&0&\frac{1}{2}&0&0&0\\ 0&0&\frac{1}{2}&0&\frac{1}{2}&0&0\\ 0&0&0&\frac{1}{2}&0&\frac{1}{2}&0\\ 0&0&0&0&\frac{1}{2}&0&1\\ 0&0&0&0&0&\frac{1}{2}&0 \end{pmatrix}\\ \] In order words, \((M_n)_{i,i+1}=(M_n)_{i+1,i}=\frac{1}{2}\) except for \((M_n)_{2,1}=(M_n)_{n1,n}=1\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1The old question has gotten so convoluted I figured that I will post a new one.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0This is just an algorithm...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0seems diagonalizable , but look at the diagonal zero which might have zero and its determinant might be zero in both and not being singular not sure though i can't help. maybe @Empty

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0find the eigen values and the eigen vectors

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1As far as I can check, from \(M_3\) to \(M_{15}\) the matrices all have distinct eigenvalues.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0ahh you are trying to show for the whole family?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Yes. Favard's theorem seems to apply in here. Those polynomials should be orthogonal.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, there is a reason why i hate orthogonality. i'll watch and learn!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 @Kainui @Empty when ever your free just lets try on this!

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I know nothing about Favard's theorem but I found this seemingly useful theorem on google.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I have no idea whether \(A_n\) is orthogonal or not. It does not seem like the case.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Can you use the Cayley Hamilton theorem? It says that every matrix satisfies its characteristic equation.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1What I want to prove is that the matrices \(M_n\) is always diagonalisable. A sufficient but not necessary condition is that the characteristic polynomials of \(M_n\) have n distinct roots. Not sure how Cayley Hamilton Theorem will help though.
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